Central configurations of nested rotated regular tetrahedra
M. Corbera
a,∗, and J. Llibre
baDepartament de Tecnologies Digitals i de la Informaci´o, Universitat de Vic, 08500 Vic, Barcelona, Catalonia, Spain
bDepartament de Matem`atiques, Universitat Aut`onoma de Barcelona, 08193 Bellaterra, Barcelona, Catalonia, Spain
Abstract
In this paper we prove that there are only two different classes of central configura- tions with convenient masses located at the vertices of two nested regular tetrahedra:
either when one of the tetrahedra is a homothecy of the other one, or when one of the tetrahedra is a homothecy followed by a rotation of Euler angles α = γ = 0 and β = π of the other one.
We also analyze the central configurations with convenient masses located at the vertices of three nested regular tetrahedra when one them is a homothecy of the other one, and the third one is a homothecy followed by a rotation of Euler angles α = γ = 0 and β = π of the other two.
In all these cases we have assumed that the masses on each tetrahedron are equal but masses on different tetrahedra could be different.
Key words: n–body problem, spatial central configurations, nested regular tetrahedra
1991 MSC: 70F10, 70F15
JGP subject classification codes: Classical mechanics
∗ Corresponding author.
Email addresses: [email protected](M. Corbera), [email protected](and J. Llibre).
1 Introduction
The equations of motion of the N–body problem in the 3–dimensional Eu- clidean space are
miq¨i = −
XN j=1, j6=i
G mimj
qi− qj
|qi− qj|3 , i = 1, . . . , N ,
where qi ∈ R3 is the position vector of the punctual mass mi in an inertial coordinate system and G is the gravitational constant which can be taken equal to one by choosing conveniently the unit of time. We fix the center of massPNi=1miqi/PNi=1mi of the system at the origin of R3N. The configuration space of the N–body problem in R3 is
E = {(q1, . . . , qN) ∈ R3N :
XN i=1
miqi = 0, qi 6= qj, for i 6= j} .
Given positive masses m1, . . . , mN a configuration (q1, . . . , qN) ∈ E is central if there exists a positive constant λ such that
¨
qi = −λ qi , i = 1, . . . , N . (1) That is if the acceleration ¨qi of each point mass mi is proportional to its position qirelative to the center of mass of the system and is directed towards the center of mass. On a central configuration system (1) can be written as
XN j=1, j6=i
mj
qi− qj
|qi− qj|3/2 = λ qi , i = 1, . . . , N . (2) Therefore a central configuration (q1, . . . , qN) ∈ E of the N–body problem with positive masses m1, . . . , mN is a solution of (2) with λ > 0.
Two central configurations in R3 are in the same class if there exists a rotation and a homothecy of R3 which transform one into the other.
A general rotation in the 3-dimensional Euclidean space having the origin of coordinates fixed is an element of SO(3) that can be parametrized by using the Euler angles (α, β, γ) via the rotation matrix R given by
cos γ sin γ 0
− sin γ cos γ 0
0 0 1
1 0 0
0 cos β sin β 0 − sin β cos β
cos α sin α 0
− sin α cos α 0
0 0 1
,
where α ∈ [0, 2π), β ∈ [0, π] and γ ∈ [0, 2π), see for more details [4].
Consider two regular tetrahedra with vertices qi and qi+4 for i = 1, . . . , 4 respectively. We say that these two tetrahedra are nested if they have the same center and the positions of the vertices of the two tetrahedra satisfy the relation qi+4= ρ R qi for all i = 1, . . . , 4, for some scale factor ρ and for some rotation matrix R. Notice that if ρ = 1, then R 6= Id where Id denotes the identity matrix.
In a similar way if we consider three regular tetrahedra with vertices qi, qi+4 and qi+8 for i = 1, . . . , 4 respectively, then these three tetrahedra are called nested if they have the same center and the positions of the vertices of the three tetrahedra satisfy the relations qi+4 = ρ R qi and qi+8 = R R′qi for all i = 1, . . . , 4, for some scale factors ρ and R and for some rotation matrices R and R′.
In [6] and in [2] it is proved that the configuration formed by 4 equal masses m1 located at the vertices of a regular tetrahedron and 4 additional masses m2 located at the vertices of a second nested regular tetrahedron is central for the 8–body problem when R = Id and the ratio of the masses m2/m1 and the scale factor ρ satisfy a convenient relation. In [5] it is proved that this kind of central configurations is also central when R is the rotation matrix of Euler angles α = 0, β = π and γ = 0. In Section 2 we summarize these results and we prove that these are the two unique classes of central configurations formed by two nested regular tetrahedra of the 8–body problem, see Theorem2.
In Section 3 we provide a new class of central configurations of the 12–body problem consisting of three nested regular tetrahedra with one of them rotated with respect to the other two by a rotation matrix with Euler angles α = 0, β = π and γ = 0. Of course on each tetrahedron the four masses located on its vertices are equal, see Result 3. The central configurations of the 12–body problem consisting of three nested regular tetrahedra when R = R′ = Id have been described in [3].
2 Two nested tetrahedra
In this section we study the spatial central configurations of the 8–body prob- lem with the masses located at the vertices of two nested regular tetrahedra.
In particular we give all classes of central configurations formed by two nested regular tetrahedra of the 8–body problem
We assume that the masses on each tetrahedron are equal but the masses on different tetrahedra could be different. Taking conveniently the unit of masses
we can assume that the masses of the tetrahedron with scale factor 1 are equal to one, that is m1 = m2 = m3 = m4 = 1, and the ones of the tetrahedron with scale factor ρ are equal to m, that is m5 = m6 = m7 = m8 = m. We also choose the unit of length in such a way that the edges of the tetrahedron with scale factor 1 have length 2.
Recall that the set of spatial central configurations is invariant under homoth- ecies and rotations of SO(3). So without loss of generality we can assume that ρ > 1 and that the positions of the vertices of the tetrahedron with scale factor 1 are a1 = (0, 0,q3/2)T, a2 = (0, 2/√
3, −1/√
6)T, a3 = (1, −1/√
3, −1/√ 6)T and a4 = (−1, −1/√
3, −1/√ 6)T.
In what follows a configuration of two nested regular tetrahedra means the configuration consisting of four equal masses m1 = m2 = m3 = m4 = 1 at the vertices of a regular tetrahedron having the positions ai for i = 1, . . . , 4 and four additional masses m5 = m6 = m7 = m8 = m at the vertices of a regular tetrahedron having positions qi+4 = ρ R ai for i = 1, . . . , 4, for some scale factor ρ > 1 and for some rotation matrix R.
We say that two configurations of two nested regular tetrahedra with ro- tation matrices R and R′ respectively are in the same class if R and R′ are such that the set {q5, q6, q7, q8} is the same for R and R′; that is, if {Ra1, Ra2, Ra3, Ra4} = {R′a1, R′a2, R′a3, R′a4}.
As usual < u, v >= uTv denotes the scalar product of the vectors u and v.
Lemma 1 Let
f (i) =
X4 j=1
3/2 − ρ < ai, Raj >
(3ρ2/2 − 2ρ < ai, Raj > +3/2)3/2, g(i) =
X4 j=1
3ρ/2− < Rai, aj >
(3ρ2/2 − 2ρ < Rai, aj > +3/2)3/2, for i = 1, . . . , 4.
A configuration of two nested regular tetrahedra is central for the 8–body prob- lem if and only if one of the following statements hold.
(a) f (1) = f (2) = f (3) = f (4), g(1) = g(2) = g(3) = g(4), 2f (1)/3 − 1/(2ρ3) 6=
0,
λ = 1/2 + 2mf (i)/3 > 0 and m = 2g(1)/(3ρ) − 1/2 2f (1)/3 − 1/(2ρ3) > 0.
(b) f (1) = f (2) = f (3) = f (4), g(1) = g(2) = g(3) = g(4), 2f (1)/3 − 1/(2ρ3) =
2g(1)/(3ρ) − 1/2 = 0 and λ = 1/2 + 2mf(i)/3 > 0 and m > 0.
PROOF. It is easy to check that at the configuration of two nested tetrahe- dra, system (2) can be written as
X4 j=1,j6=i
ai− aj
|ai− aj|3 + m
X4 j=1
ai − ρRaj
|ai− ρRaj|3 = λai , (3)
X4 j=1
ρRai− aj
|ρRai− aj|3 + m
X4 j=1,j6=i
ρRai− ρRaj
|ρRai − ρRaj|3 = λρRai , (4) for i = 1, . . . , 4. We do the scalar product of the i–th equation of (3) with the vector ai and the scalar product of the i–th equation of (4) with the vector Rai and we get
X4 j=1,j6=i
< ai, ai > − < ai, aj >
|ai− aj|3 + m
X4 j=1
< ai, ai > −ρ < ai, Raj >
|ai− ρRaj|3
= λ < ai, ai > , (5)
X4 j=1
ρ < Rai, Rai > − < Rai, aj >
|ρRai− aj|3 + m
X4 j=1,j6=i
ρ < Rai, Rai > −ρ < Rai, Raj >
|ρRai− ρRaj|3
= λρ < Rai, Rai > , (6)
for i = 1, . . . , 4. It is easy to check that < ai, ai >=< Rai, Rai >= 3/2,
< Rai, Raj >=< ai, aj >, |ρRai − ρRaj|3 = ρ3|ai− aj|3 and
X4 j=1,j6=i
< ai, ai > − < ai, aj >
|ai− aj|3 = 3 4 , for all i = 1, . . . , 4. On the other hand
|ai− ρRaj| =3ρ2/2 − 2ρ < ai, Raj > +3/21/2,
|ρRai− aj| =3ρ2/2 − 2ρ < Rai, aj > +3/21/2 . So equations (5) and (6) become
3
4+ m f (i) = 3
2λ , g(i) + m 3
4ρ2 = 3ρ
2 λ , (7)
for i = 1, . . . , 4. In short, the configuration consisting of two nested tetrahedra is central if and only if λ and m is a solution of (7) satisfying that λ > 0 and m > 0.
From the first equation of (7) we have that λ = 1/2+2mf (i)/3 for i = 1, . . . , 4, this implies that f (1) = f (2) = f (3) = f (4). From the second equation of (7) we have λ = 2g(i)/(3ρ) + m/(2ρ3) for i = 1, . . . , 4, so g(1) = g(2) = g(3) = g(4). Equating both expressions of λ for i = 1 we have that if 2f (1)/3 − 1/(2ρ3) 6= 0, then
m = 2g(1)/(3ρ) − 1/2 2f (1)/3 − 1/(2ρ3). This proves statement (a).
Moreover if 2f (1)/3 − 1/(2ρ3) = 2g(1)/(3ρ) − 1/2 = 0, then the solution of (7) is λ = 1/2 + 2mf (i)/3 > 0 and m > 0, which concludes the prove of statement (b).
Theorem 2 The following statements hold.
(a) There are two unique classes of central configurations of two nested regular tetrahedra.
(a.1) The class of configurations of Type I with {q5, q6, q7, q8} = {ρa1, ρa2, ρa3, ρa4}.
(a.2) The class of configurations of Type II with {q5, q6, q7, q8} = {ρRa1, ρRa2, ρRa3, ρRa4} and
R = P =
1 0 0
0 −1 0 0 0 −1
.
These two classes of configurations are shown in Figure 1.
(b) (see [2,6]) The configuration of Type I is central for the spatial 8–body prob- lem when
m = mI(ρ) = n(ρ) d(ρ) =
(2/3)3/2 (ρ − 1)2 − ρ
2 + 2√
2 (3ρ + 1) (3ρ2+ 2ρ + 3)3/2
−1/2
ρ2 − (2/3)3/2ρ
(ρ − 1)2 + 2√
2 ρ (ρ + 3) (3ρ2+ 2ρ + 3)3/2
,
and ρ > α = 1.8899915758 . . . , where α is the unique real solution of n(ρ) = 0 for ρ > 1. Moreover, fixed a value of m > 0 there exists a unique ρ > α for which the configurations of Type I is central.
(c) (see [5]) The configuration of Type II is central for the spatial 8–body problem when
m = mII(ρ) = n(ρ) d(ρ) =
−ρ
2+ 2√
2(3ρ − 1)
(3ρ2− 2ρ + 3)3/2 + (2/3)3/2 (ρ + 1)2
− 2√
2(ρ − 3)ρ
(3ρ2− 2ρ + 3)3/2 + (2/3)3/2ρ (ρ + 1)2 − 1
2ρ2 ,
(a) Conf. of Type I ρ > 1
(b) Conf. of Type II 1 6 ρ 6 3
(c) Conf. of Type II ρ > 3
Fig. 1. Plot of the possible classes of configurations of two nested regular tetrahedra.
Notice that we have two possibilities for the configurations of Type II, in (b) the faces of both tetrahedra intersect whereas in (c) they do not intersect.
and ρ ∈ [1, α1) ∪ (α2, ∞) where α1 = 1.3981650369 . . . is the unique real solution of n(ρ) = 0 with ρ > 1 and α2 = 6.5360793703 . . . is the unique real solution of d(ρ) = 0 with ρ > 1. Moreover the following statements hold.
(c.1) There is a unique central configuration of Type II with ρ ∈ [1, α1) for m ∈ (0, 1]. In this configuration the faces of both tetrahedra intersect.
(c.2) Let α3 = 8.7756058918 . . . be the unique real solution of (n(ρ)/d(ρ))′ with ρ > 1 and m0 = n(α3)/d(α3) = 2880.33 . . . . There are no central configu- rations for m ∈ (1, m0).
(c.3) There is a unique central configuration of Type II with ρ = α3 for m = m0. In this configuration the faces of the tetrahedra do not intersect.
(c.4) There are two central configurations of Type II, one with ρ ∈ (α2, α3) and the other with ρ ∈ (α3, ∞), for m ∈ (m0, ∞). In this configuration the faces of the tetrahedra do not intersect.
In [6,5] the authors also consider values of the scale factor ρ ∈ (0, 1]. The results for the values of ρ in (0, 1] can be obtained from the ones given in Theorem 2 by replacing m by 1/m and ρ by 1/ρ. In particular, fixed a value of m > 0 there exists a unique 0 < ρ < 1/α = 0.5291028874 . . . for which the configurations of Type I is central. There is a unique central configuration of Type II with ρ ∈ (1/α1, 1] = (0.7152231486 . . . , 1] for m ∈ [1, +∞). There is a unique central configuration of Type II with ρ = 1/α3 = 0.1139522458 . . . for m = 1/m0 =fm0 = 0.0003471823 . . . . There are two central configurations of Type II, one with ρ ∈ (1/α3, 1/α2) = (0.1139522458 . . . , 0.1529969280 . . . ) and the other one with ρ ∈ (0, 1/α3) = (0, 0.1139522458 . . . ), for m ∈ (0,fm0).
PROOF. [Proof of Theorem 2] We know that the spatial central configura-
tions are invariant under rotations of SO(3). Consider the rotation
A =
1/2 √
3/2 0
1/(2√
3) −1/6 2√
q 2/3
2/3 −√
2/3 −1/3
.
It is easy to check that Aa1 = a2, Aa2 = a3, Aa3 = a1 and Aa4 = a4. Since the rotation A leaves invariant the tetrahedron with scale factor 1, it also must leave invariant the one with scale factor ρ. This means that the sets {q5, q6, q7, q8} and {Aq5, Aq6, Aq7, Aq8} must be the same. Then either Aq5 = q5, or Aq5 = q6, or Aq5 = q7, or Aq5 = q8. On the other hand, by Lemma1the rotation matrix R provides a central configuration of two nested tetrahedra if f (1) = f (2) = f (3) = f (4) and g(1) = g(2) = g(3) = g(4).
We start analyzing the solutions of system f5 = Aq5− q5 = 0 where
f5 =
−sin β√
3 sin γ − 3 cos γ 2√
2 cos β 2
√3 + 1
12sin β3√
2 sin γ − 7√
6 cos γ sin β sin γ − cos γ
√3
!
− 2
s2 3cos β
.
From the first equation of f5 = 0 we have that either β = 0, or β = π, or γ = π/3, or γ = 4π/3. It is easy to check that β = 0 and β = π do not provide solutions of the full system f5 = 0. If γ = π/3, then there is a unique solution of f5 = 0 with β ∈ [0, π], the solution (α, β, γ) = (α, 2 arctan(1/√
2), π/3) = σ1(α). When γ = 4π/3 system f5 = 0 has a unique solution with β ∈ [0, π]
which is given by (α, β, γ) = (α, 2 arctan√
2, 4/3) = σ2(α). These two solu- tions of f5 = 0 leave invariant the tetrahedron with scale factor ρ.
It only remains to find the values of α for which σ1(α) and σ2(α) satisfy that f (1) = f (2) = f (3) = f (4) and g(1) = g(2) = g(3) = g(4). The functions f (i) and g(i) evaluated at σ1(α) are given by
f (1) = f (2) = f (3) = −
√2(ρ − 3)
(3ρ2− 2ρ + 3)3/2 + F (α) + F (α + 2π/3) + F (α + 4π/3), f (4) =
q2/3
(ρ + 1)2 − 3√
2(ρ − 3) (3ρ2− 2ρ + 3)3/2, g(1) = 3√
2(3ρ − 1) (3ρ2− 2ρ + 3)3/2 +
q2/3 (ρ + 1)2, g(2) = g(3) = g(4) =
√2(3ρ − 1)
(3ρ2− 2ρ + 3)3/2 + G(α) + G(α + 2π/3) + G(α + 4π/3),
where
F (α) =
√6(8ρ cos α + ρ + 9) (9ρ2+ 16ρ cos α + 2ρ + 9)3/2, G(α) =
√6(9ρ + 8 cos α + 1) (9ρ2+ 16ρ cos α + 2ρ + 9)3/2.
We must find the values of α for which σ1(α) is a solution of system e1(α) = f (1) − f(4) = 0 and e2(α) = g(1) − g(2) = 0. We consider the additional equation
e(α) = e1(α) + e2(α)
ρ − 1 = H0− H(α) − H(α + 2π/3) − H(α + 4/π/3) = 0, where
H0= 8√
2
(3ρ2− 2ρ + 3)3/2,
H(α) = 8√
6(1 − cos α)
(9ρ2+ 16ρ cos α + 2ρ + 9)3/2.
Notice that H0 > 0 for all ρ > 1 and H(α) > 0 for all ρ > 1 and α ∈ [0, 2π).
Next we prove that e(α) 6 0 for all ρ > 1 and α ∈ [0, 2π). First we find the minimum values of the function h(α) = H(α) + H(α + 2π/3) + H(α + 4/π/3).
It is easy to check that h(α) is a periodic function of period 2π/3. Moreover h(2π/3 − α) = h(α) so it is sufficient to study the behaviour of h(α) for α ∈ [0, π/3]. The possible minimums of h(α) are given by the solutions of h′(α) = 0 where
h′(α) = H′(α) + H′(α + 2π/3) + H′(α + 4π/3), and
H′(α) = 8√
6 (9ρ2− 8ρ cos α + 26ρ + 9) sin α (9ρ2+ 16ρ cos α + 2ρ + 9)5/2 .
In order to solve equation h′(α) = 0, we eliminate the radicals and the de- nominators of the equation by using the procedure described in the Appendix and we set c = cos α. The result is the polynomial equation
21233664(c − 1)(c + 1)(2c − 1)2(2c + 1)2ρ2E(c, ρ) = 0, (8) in the variable c with coefficients depending on ρ where E(c, ρ) is a polynomial of degree 21 in c. Notice that we are only interested in solutions of h′(α) = 0 with α ∈ [0, π/3], so we only consider solutions of (8) with c ∈ [1/2, 1].
We solve symbolically with the help of the algebraic manipulator Mathematica the equation E(c, ρ) = 0. We see that only three of the 21 solutions c(ρ) are real for all ρ > 1, one satisfies c(ρ) > 1.8638527724 . . . for all ρ > 1, the other one satisfies c(ρ) > 1.9150397104 . . . , and finally the third one c(ρ) = ˜c(ρ) satisfies ˜c(ρ) ∈ [0.5088804885 . . . , 1] for ρ ∈ [1, ρ∗] and ˜c(ρ) > 1 for ρ > ρ∗. Here ρ∗ = 6.1180170733 . . . is the unique real solution of E(1, ρ) = 0 with ρ > 1. Two of the solutions of E(c, ρ) = 0 are real for ρ ∈ [1, ρ∗] and in C \ R for ρ > ρ∗, moreover for these two real solutions c(ρ) 6∈ [1/2, 1]. The rest of the solutions of E(c, ρ) = 0 are in C \ R for all ρ > 1. In short, there is a unique solution c(ρ) = ˜c(ρ) with ρ ∈ [1, ρ∗] satisfying that c(ρ) ∈ [1/2, 1]. This solution provides a solution of the initial equation h′(α) = 0 only when ρ = ρ∗ and c = 1. Consequently the solutions of (8) with c(ρ) ∈ [1/2, 1] that provide solutions of h′(α) = 0 are c = 1/2 and c = 1 for all ρ > 1. Therefore h′(α) = 0 has two unique solutions in the interval [0, π/3] that are independent of ρ, α = 0 which corresponds to a minimum and α = π/3 which corresponds to a maximum. Thus the minimum values of h(α) in the interval [0, 2π) are α = 0, α = 2π/3 and α = 4π/3. Furthermore h(0) = h(2π/3) = h(4π/3) = H0. Consequently e(α) 6 0 for all ρ > 1 and e(α) = 0 if and only if either α = 0, α = 2π/3 or α = 4π/3. It is easy to check that all these solutions satisfy system e1(α) = 0 and e2(α) = 0. On the other hand, in a similar way it can be proved that the solutions of system e1(α) = 0 and e2(α) = 0 for ρ = 1 are also α = 0, α = 2π/3 or α = 4π/3.
In short, there are three rotations for σ1(α) that satisfy conditions of Lemma1, they have Euler rotation angles (α, β, γ) equal to
(0, 2 arctan(1/√
2), π/3), (2π/3, 2 arctan(1/√
2), π/3), (4π/3, 2 arctan(1/√
2), π/3).
All these rotations give configurations of Type II.
Proceeding in a similar way with the solution σ2(α) we obtain three additional rotations that satisfy conditions of Lemma1, they have Euler rotation angles (α, β, γ) equal to
(π/3, 2 arctan√
2, 4π/3), (π, 2 arctan√
2, 4π/3), (5π/3, 2 arctan√
2, 4π/3).
All these rotations give configurations of Type I.
We have just proved that all the rotations satisfying f5 = 0 and the conditions of Lemma1 provide configurations of Type either I or II.
Type I Type II
(α, 0, 2π − α) with α ∈ [0, 2π) (α, π, α) with α ∈ [0, 2π) (π, 2 arctan√
2, 2π/3) (2π/3, 2 arctan(1/√ 2), π) (π/3, 2 arctan√
2, 0) (0, 2 arctan(1/√
2), 5π/3) Table 1
The Euler rotation angles (α, β, γ) of all the rotations that satisfy system Aq5−q6 = 0 and the conditions of Lemma1 classified by the type of configurations that they provide.
Type I Type II
(α, 0, 4π/3 − α mod 2π), α ∈ [0, 2π) (α, π, α + 2π/3 mod 2π), α ∈ [0, 2π) (π/3, 2 arctan√
2, 2π/3) (4π/3, 2 arctan(1/√
2), 5π/3) (5π/3, 2 arctan√
2, 0) (0, 2 arctan(1/√
2), π) Table 2
The Euler rotation angles (α, β, γ) of all the rotations that satisfy system Aq5−q7 = 0 and the conditions of Lemma1 classified by the type of configurations that they provide.
Type I Type II
(α, 0, 2π/3 − α mod 2π), α ∈ [0, 2π) (α, π, α − 2π/3 mod 2π), α ∈ [0, 2π) (π, 2 arctan√
2, 0) (2π/3, 2 arctan(1/√
2), 5π/3) (5π/3, 2 arctan√
2, 2π/3) (4π/3, 2 arctan(1/√ 2), π) Table 3
The Euler rotation angles (α, β, γ) of all the rotations that satisfy system Aq5−q8 = 0 and the conditions of Lemma1 classified by the type of configurations that they provide.
By using similar arguments, we find all the rotations that satisfy system Aq5− qi = 0, for i = 6, 7, 8, and the conditions of Lemma1, moreover we see that all these rotations provide configurations of Type either I or II. The Euler rotation angles (α, β, γ) of those rotations classified by the type of configurations that they provide are given in Tables 1, 2, and 3.
In short, all rotations that satisfy conditions of Lemma 1 provide configura- tions of Type either I or II. This completes the proof of statement (a).
Statement (b) is proved in [2] by using the choice of the units of mass and length that we use here and it is proved in [6] with a different choice of the units. Here by using Lemma 1 we give a new and shorter proof of this result.
Indeed, statement (b) is an immediate consequence of Lemma1together with statement (a). In statement (a) we have proved that if q5 = ρa1, q6 = ρa2, q7 = ρa3, and q8 = ρa4 then the hypotheses of Lemma 1 hold. So applying Lemma 1 we obtain the expressions of λ and m, these expressions provide a
central configuration if λ > 0 and m > 0. We see that 2f (1)/3 − 1/(2ρ3) 6= 0 for all ρ > 1 and 2g(1)/(3ρ) − 1/2 = 0 when ρ = α, in particular, λ > 0 and m > 0 when ρ > α. Moreover m is an increasing function of ρ, so for each value of m > 0 there exists a unique value of ρ > α for which the configuration is central. This completes the proof of the statement (b).
Statement (c) is proved in [5] by using different choices of the units of mass and length that the ones used here. Statement (c) is also an immediate consequence of Lemma 1 together with statement (a). As above in statement (a) we have proved that if q5 = ρP a1, q6 = ρP a2, q7 = ρP a3, and q8 = ρP a4 then the hypotheses of Lemma 1 hold. So applying this Lemma 1 we obtain the expressions of λ and m which provide a central configuration if λ > 0 and m > 0. We see that 2f (1)/3 − 1/(2ρ3) = 0 at ρ = α2 and 2g(1)/(3ρ) − 1/2 = 0 at ρ = α1, in particular, λ > 0 and m > 0 when ρ ∈ [1, α1) ∪ (α2, ∞).
Moreover m = m(ρ) is decreasing at the interval [1, α2) with m(1) = 1 and limρ→α−
2 m(ρ) = −∞, it is decreasing at the interval (α2, α3), increasing at the interval (α3, ∞), the point ρ = α3 is a minimum with m(ρ) = m0 and limρ→α+
2 m(ρ) = ∞ and limρ→+∞m(ρ) = ∞. These properties of m(ρ) together with the fact the faces of the nested tetrahedra intersect when 1 6 ρ 6 3 prove statements (c.1), (c.2), (c.3), and (c.4).
3 Three nested tetrahedra
In this section we study the spatial central configurations of the 12–body problem when the masses are located at the vertices of three nested regular tetrahedra with scale factors 1, ρ and R and some of them rotated with respect to the others by a rotation of Euler angles α = 0, β = π and γ = 0 (i.e. by a rotation with rotation matrix P ). Taking conveniently the unit of masses we can assume that all the masses of the tetrahedron with scale factor 1 are equal to one. We also choose the unit of length in such a way that the edges of the tetrahedron with scale factor 1 have length 2. Without loss of generality we can assume that the configuration has only one rotated tetrahedra which could be either the inner, the medium or the outer one. This is due to the fact that central configurations are invariant under rotations and a configuration with two rotated tetrahedra can be transformed into a configuration with only one rotated tetrahedra by doing a rotation of Euler angles α = 0, β = π and γ = 0. Central configurations are also invariant under homothecies, so we can assume that the rotated tetrahedra is the one with scale factor ρ, and that 1 6 ρ < R when the rotated tetrahedra is the medium one, 1 < R 6 ρ when the rotated tetrahedra is the outer one, and finally 0 < ρ < 1 < R when the rotated tetrahedra is the inner one (see Figure 2). We define
(a) 1 6 ρ < R (b) 1 < R 6 ρ (c) 0 < ρ < 1 < R
Fig. 2. Three possible central configurations with three nested regular tetrahedra with scale factors 1, ρ and R and the one with scale factor ρ rotated with respect to the other two by a rotation of Euler angles α = 0, β = π and γ = 0. In (a) the medium tetrahedra is the rotated one, in (b) is the outermost and in (c) is the innermost. In any of these three possibilities the tetrahedra can intersect or not.
Cm= {(ρ, R) ∈ R2 : 1 6 ρ < R}, Co= {(ρ, R) ∈ R2 : 1 < R 6 ρ},
Ci= {(ρ, R) ∈ R2 : 0 < ρ < 1 < R}.
Result 3 Consider four equal masses m1 = m2 = m3 = m4 = 1 at the vertices of a regular tetrahedron with edge length 2 having positions a1, a2, a3
and a4. Consider four additional equal masses m5 = m6 = m7 = m8 = m at the vertices of a second nested regular tetrahedron having positions qi+4= ρP ai for all i = 1, . . . , 4 with ρ > 0. Finally we consider four additional equal masses m9 = m10 = m11 = m12 = M at the vertices of a third nested regular tetrahedron having positions qi+8 = Rai for all i = 1, . . . , 4 with R > 1 (see Figure 2). Then the following statements hold.
(a) Such configuration is central for the spatial 12–body problem when m = m(ρ, R) and M = M(ρ, R) are given by the expression (11) and (R, ρ) ∈ D = {(R, ρ) ∈ R2 : det(A) 6= 0, m(R, ρ) > 0, M(R, ρ) > 0, ρ > 0, R > 1}
(see (11) for the definition of A and see Figure 3 for the plot of D).
(b) The set D is formed by the disjoint union of the sets Di for i = 1, . . . , 6 defined in Figure 3. Then the regions Di provide central configurations of the 12–body problem with the masses located at the vertices of three nested regular tetrahedra satisfying the following.
(b.1) The rotated tetrahedra is the inner one and it intersects only the medium when (ρ, R) ∈ D2∩Ci, it intersects the medium and the outer when (ρ, R) ∈ D4∩ Ci, and the three tetrahedra do not intersect when (ρ, R) ∈ D1. (b.2) The rotated tetrahedra is the medium one and it intersects only the inner
when (ρ, R) ∈ D2∩ Cm and when (ρ, R) ∈ D3∩ {ρ 6 3}, it intersects the inner and the outer when (ρ, R) ∈ D5∩ Cm ∩ {ρ 6 3} and when (ρ, R) ∈ D4∩ Cm, it intersects only the outer one when (ρ, R) ∈ D5∩ Cm∩ {ρ > 3}, and the three tetrahedra do not intersect when (ρ, R) ∈ D3∩ {ρ > 3}.
R R
ρ ρ
p q
D2 D3
D1
D1
D4
D5
R
ρ
D5
D6
Fig. 3. The set D. The dotted curves correspond to points where det(A) = 0, the continuous curves correspond to points where m(ρ, R) = 0, and finally, the curves with small circles correspond to points where M (ρ, R) = 0.
(b.3) The rotated tetrahedra is the outer one and it intersects only the medium when (ρ, R) ∈ D5∩ Co∩ {ρ > 3}, it intersects the inner and the medium when (ρ, R) ∈ D5∩ Co∩ {ρ 6 3}, and the three tetrahedra do not intersect when (ρ, R) ∈ D6.
(c) Let
m(M, ρ, R) = ρ2(ρ + 2f (ρ, 1))
2ρ3f (1, ρ) + 1 − M2ρ2(ρg1(R) − f(ρ, R)) 2ρ3f (1, ρ) + 1 ,
and M0(ρ, R) = (ρ2(ρ+2f (ρ, 1)))/(2ρ2(ρg1(R)−f(ρ, R)) (see (9) for the def- initions of f and g1), and let p = (p1, p2) = (1.5094116757 . . . , 1.9479968088 . . . ) and q = (q1, q2) = (1.9926247501 . . . , 31.4606148079 . . . ) be the solutions of system det(A) = det(A2) = det(A3) = 0 (see (11) for the definitions of A2
and A3). Then the following statements hold.
(c.1) For each M > M0(p) = 0.1252891302 . . . and m = m(M, p) = −0.4001317738 · · ·+
3.1936671053 . . . M we have a central configuration of the spatial 12–body problem when (ρ, R) = p. In this configuration the rotated tetrahedra is the medium and it intersects with the inner and the outer.
(c.2) For each M > M0(q) = 25204.620455 . . . and m = m(M, q) = −3.5811747685 · · ·+
0.0001420840585 . . . M we have a central configuration of the spatial 12–
body problem when (ρ, R) = q. In this configuration the rotated tetrahedra is the medium and it intersects only the inner.
The values of the masses m(ρ, R) and M(ρ, R) that provide central configu- rations on sets Di are analyzed in Subsection 3.1.
We call the previous result as Result 3 instead of Theorem 3 because part of its proof is done numerically with the help of Mathematica.
PROOF. [Proof of Result3] The positions and the values of the masses have been taken so that the center of mass of the configuration is located at the origin of coordinates.
We substitute the positions and the values of the masses into (2). After some computations we obtain that system (2) over the nested three tetrahedra con- figuration with R > 1 is equivalent to system Ax = b, more precisely
1 f (1, ρ) g1(R) ρ −1/(2ρ2) f (ρ, R) R f (R, ρ) −1/(2R2)
λ m M
=
1/2
−f(ρ, 1) g2(R)
, (9)
where
f (x, y) = 2√
2(y − 3x)
(3x2− 2yx + 3y2)3/2 − (2/3)3/2 (x + y)2 , g1(x) = − 2√
2(x + 3)
(3x2+ 2x + 3)3/2 + (2/3)3/2 (x − 1)2 , g2(x) = 2√
2(3x + 1)
(3x2+ 2x + 3)3/2 + (2/3)3/2 (x − 1)2 .
Since 3x2− 2yx + 3y2 > 0 in D = {(x, y) ∈ R2 : x > 0, y > 0}, the function f (x, y) is defined for all (x, y) ∈ D. The functions g1(x) and g2(x) are defined for all x > 1, moreover
x→1lim+g1(x) = lim
x→1+g2(x) = +∞. (10)
On the other hand f (x, y) < 0 for all (x, y) ∈ D. Indeed the set D can be written as D = {(x, y) ∈ R2 : y = a x, x > 0, a > 0}. It is easy to check that f (x, a x) = f1(a)/x2 where
f1(a) = 2√
2(a − 3)
(3a2− 2a + 3)3/2 − (2/3)3/2 (a + 1)2 .
As in the Appendix we find the zeroes of f1(a) by reducing the problem to find the zeroes of a polynomial of one variable and we see that f1(a) = 0 has no real solutions for a > 0. In particular f1(a) < 0 for a > 0.
If det(A) 6= 0, then the solution of system (9) is
λ = λ(ρ, R) = det(A1)
det(A) = 1 det(A)
1/2 f (1, ρ) g1(R)
−f(ρ, 1) −1/(2ρ2) f (ρ, R) g2(R) f (R, ρ) −1/(2R2)
,
m = m(ρ, R) = det(A2)
det(A) = 1 det(A)
1 1/2 g1(R) ρ −f(ρ, 1) f(ρ, R) R g2(R) −1/(2R2)
, (11)
M = M(ρ, R) = det(A3)
det(A) = 1 det(A)
1 f (1, ρ) 1/2 ρ −1/(2ρ2) −f(ρ, 1) R f (R, ρ) g2(R)
.
The solution λ(ρ, R), m(ρ, R) and M(ρ, R) gives a central configuration of the 12–body problem if and only if R and ρ are such that λ(ρ, R) > 0, m(ρ, R) > 0 and M(ρ, R) > 0.
It is easy to check that
det(A) = ρ
2R2f (1, ρ) + 1
4R2ρ2 + Rf (ρ, R)f (1, ρ) − f(R, ρ)f(ρ, R) + R
2ρ2g1(R) + ρf (R, ρ) g1(R) .
We analyze the sign of det(A) on the boundaries of the regions Cm, Co and Ci. Clearly
ρ→0lim+det(A) = sign
1 4R2 + R
2 g1(R)
· ∞ , and from (10)
R→1lim+det(A) = sign 1
2ρ2 + ρf (1, ρ)
!
· ∞ .
We solve equation f2(R) = 1/(4R2)+Rg1(R)/2 = 0 as in the Appendix and we see that it has no real solutions for R > 1, moreover f2(R) > 0 for all R > 1.
Solving equation f3(ρ) = 1/(2ρ2)+ρf (1, ρ) = 0 as in the Appendix we find two
unique real roots ρ = b1 = 0.7152231486 . . . and ρ = b2 = 6.5360789462 . . . . Moreover f3(ρ) > 0 for ρ ∈ (0, b1) ∪ (b2, ∞) and f3(ρ) < 0 for ρ ∈ (b1, b2).
In short, limρ→0+det(A) = +∞ for all R > 1, limR→1+det(A) = +∞ when ρ ∈ (0, b1) ∪ (b2, ∞) and limR→1+det(A) = −∞ when ρ ∈ (b1, b2). The regions of these two limits are indicated in Figure 4.
After some computations we see that when ρ = 1 and R > 1
det(A) = f4(R) = −−9 + 9√ 2 +√
6 36R2 + 3√
6R (R2+ 2R + 1) − 8 27(R − 1)2(R + 1)2 +23√
3R(R + 1)2+ R(R + 1)2− 4
27(R + 1)4 +8 (3R2− 10R + 3) (3R2− 2R + 3)3
−29 +√
3(R − 3)R
9 (3R2− 2R + 3)3/2 − 8 (5R2− 2R + 1) 3√
3(R − 1)2(R + 1) (3R2 − 2R + 3)3/2
−(R + 3)9√
2R3+ 18√
2R2+ 9√
2R − 8√ 3 9(R + 1)2(3R2+ 2R + 3)3/2
+ 8 (3R2+ 8R − 3)
(3R2 − 2R + 3)3/2(3R2+ 2R + 3)3/2.
We solve equation f4(R) = 0 as in the Appendix and we find two real so- lutions with R > 1, they are R = c1 = 1.3711124500 . . . and R = c2 = 28.2412927769 . . . . Moreover f4(R) > 0 for R ∈ (c1, c2) and f4(R) < 0 for R ∈ (1, c1) ∪ (c2, ∞).
For R = ρ and ρ > 1 we have that
det(A) = f5(ρ) =
−18 + 9√
2 − 2√
3(ρ − 3)
9ρ (3ρ2− 2ρ + 3)3/2 +−2 − 6√
3 + 3√ 6 27(ρ − 1)2ρ +
18 − 9√
2 + 2√
3(ρ + 3)
9ρ (3ρ2+ 2ρ + 3)3/2 + 2 + 6√
3 − 3√ 6
27ρ(ρ + 1)2 − 29 + 12√ 3 108ρ4 . We solve equation f5(ρ) = 0 as in the Appendix and we see that it has no real solutions with ρ > 1. Moreover f5(R) < 0 for all ρ > 1.
Notice that det(A) is continuous for all ρ > 0, R > 1. This means that if det(A) has different signs on two different curves γ1 and γ2, then there exist an odd number of curves of zeroes of det(A) (taking into account their multiplicities) between γ1 and γ2, whereas if det(A) has the same sign on the curves γ1 and γ2, then there exist an even number (that could be 0) of curves of zeroes of det(A) (taking into account their multiplicities) between γ1 and γ2. Moreover these curves are continuous.
0 1 2 3 4 5 23
24 25 26 27 28 29 30
0 1 2 3 4 5
1 2 3 4 5
0 5 10 15 20 25 30
5 10 15 20 25 30
R R
ρ ρ +
+ +
+ −
−
+ − − (1, c1)
(1, c2)
(b1, 1) Ci
Γ2
Γ1
Cm
Ci Cm
Co
Cm
Co Ci
Γ1
Γ2
Γ3
(b2, 1)
− + +
+ R −
ρ
Fig. 4. The level curves det(A) = 0. The dashed lines correspond to the boundaries of the regions Cm, Co and Ci.
By analyzing the sign of det(A) on the boundary of Ci we have seen that det(A) > 0 on {(ρ, R) ∈ R2 : ρ = 0, R > 1} ∪ {(ρ, R) ∈ R2 : ρ ∈ (0, b1), R = 1} ∪ {(ρ, R) ∈ R2 : ρ = 1, R ∈ (c1, c2)}, det(A) < 0 on {(ρ, R) ∈ R2 : ρ = 1, R ∈ (1, c1)} ∪ {(ρ, R) ∈ R2 : ρ = 1, R ∈ (c2, +∞)} and that det(A) = 0 at the points (1, c1) and (1, c2). Therefore on Ci there exists a curve Γ1 starting at the point (b1, 1) and passing through the point (1, c1) and another one Γ2 coming from R = +∞ and passing through the point (1, c2) on which det(A) = 0 (see Figure 4).
On the boundary of Cmwe have that det(A) < 0 everywhere with the exception of the set {(ρ, R) ∈ R2 : ρ = 1, R ∈ (c1, c2)} and det(A) = 0 at the points (1, c1) and (1, c2). Therefore we can guarantee the existence of a curve in Cm with R ∈ (c1, c2) on which det(A) = 0. This curve starts at the point (1, c1), so it is the continuation on the region Cm of the curve Γ1. By plotting the level curves det(A) = 0 with the help of Mathematica (see Figure 4), we see that the curve Γ2 can be continued to the region Cm and that the curves Γ1 and Γ2
do not coincide.
Finally on the boundary of Co we have that det(A) < 0 at everywhere with the exception of the set {(ρ, R) ∈ R2 : ρ ∈ (b2, +∞), R = 1} and det(A) = 0 at the point (b2, 1). Therefore there exists a curve Γ3 on Co starting at the point (b2, 1) and going to ρ = +∞ (see again Figure 4).
By plotting the level curves det(A) = 0 for larger values of ρ and R we see that these are the unique three curves with det(A) = 0.
Boundary Sign of det(A2) ρ > 0, R = 1 limR→1+det(A2) = +∞
ρ = 0, R > 1 det(A2) > 0
ρ = 1, R > 1 det(A2) > 0 when R ∈ (1, c3) det(A2) < 0 when R ∈ (c3, ∞) R = ρ, ρ > 1 det(A2) > 0 when ρ ∈ (1, d1)
det(A2) < 0 when ρ ∈ (d1, ∞) Table 4
The sign of det(A2) on the boundaries of Cm, Coand Ci. Here c3 = 1.7097032687 . . . is the unique solution of equation det(A2)|ρ=1= 0 with R > 1, d1= 2.4741715435 . . . is the unique solution of equation det(A2)|R=ρ = 0 with ρ > 1.
0 1 2 3 4 5
1 2 3 4 5
R
ρ
+ −
+ +
+
−
(1, c3)
(d1, d1) Ci
Cm
Co
Cm
Co
Ci
+ − −
+ R
ρ
Fig. 5. The level curves det(A2) = 0. The dashed lines correspond to the boundaries of the regions Cm, Co and Ci.
Now we analyze the sign of det(A2) on the boundaries of Cm, Co and Ci. From (11) we have that
det(A2) = g1(R)g2(R)ρ + ρ
4R2 +f (ρ, 1) 2R2 + 1
2Rf (ρ, R) + Rf (ρ, 1)g1(R)
−f(ρ, R)g2(R) .
Since the procedure is exactly the same than for the sign of det(A) we omit the computations and we only summarize the results in Table 4.
The plot of the level curves det(A2) = 0 is given in Figure5. As in the previous case the plot of these level curves for larger values of ρ and R does not provide new curves with det(A2) = 0.
Boundary Sign of det(A3)
ρ > 0, R = 1 limR→1+det(A3) = −∞ when ρ ∈ (1, b1) ∪ (b2, ∞) limR→1+det(A3) = +∞ when ρ ∈ (b1, b2)
ρ = 0, R > 1 limρ→0+det(A3) = −∞ when R ∈ (1, e1) limρ→0+det(A3) = +∞ when R ∈ (e1, ∞) ρ = 1, R > 1 det(A3) > 0 when R ∈ (1, c4)
det(A3) < 0 when R ∈ (c4, ∞) R = ρ, ρ > 1 det(A3) > 0 when ρ ∈ (1, d2)
det(A3) < 0 when ρ ∈ (d2, ∞) Table 5
The sign of det(A3) on the boundaries of Cm, Co and Ci. Here e1= 1.8899915758 . . . is the unique real solution of equation R/4 − g2(R)/2 = 0 with R > 1, the values b1
and b2defined above are the unique real solutions of equation −ρf(1, ρ)−1/(2ρ2) = 0 with ρ > 1, c4 = 1.6436467629 . . . is the unique real solution of det(A3)|ρ=1 = 0 with R > 1, and d2 = 2.0151307882 . . . is the unique real solution of equation
det(A3)|R=ρ= 0 with ρ > 1.
Finally we analyze the sign of det(A3) on the boundaries of Cm, Co and Ci. From (11) we have that
det(A3) = −f(1, ρ)f(ρ, 1)R + R 4ρ2 + 1
2ρf (R, ρ) + f (R, ρ)f (ρ, 1) − g2(R) 2ρ2
−ρf(1, ρ)g2(R) .
As above, and since the procedure is exactly the same than for the sign of det(A) we omit the computations and we only summarize the results in Ta- ble 5.
The plot of the level curves det(A3) = 0 is given in Figure6. As in the previous cases, the plot of these level curves for larger values of ρ and R does not provide new curves with det(A3) = 0.
Since f (R, ρ) < 0 and g2(R) > 0 in the region ρ > 0, R > 1, from the third equation of (9) we have that if m > 0 and M > 0, then λ > 0. So it is not necessary to find the set of level curves det(A1) = 0.
Analyzing the sign of m and M on the regions Cm, Co, and Ci we prove that the region D = {(R, ρ) ∈ R2 : det(A) 6= 0, m(R, ρ) > 0, M(R, ρ) > 0, ρ >
0, R > 1} where the solution of (9) given by (11) gives a central configuration of the spatial 12–body problem is the one plotted in Figure 2. This proves statement (a).
It is easy to check that the faces of the rotated tetrahedron, which has scale factor ρ, intersect with the ones of the tetrahedron with scale factor 1 when ρ ∈ [1/3, 3], and they intersect with the ones of the tetrahedron with scale factor R when R ∈ [ρ/3, 3ρ]. Moreover the rotated tetrahedron can be either
0 1 2 3 4 5 1
2 3 4 5
0 5 10 15 20 25 30
5 10 15 20 25 30
R
ρ +
− − +
+ +
−
−
(1, c4)
(d2, d2)
(b1, 1) (0, e1)
Ci
Cm
Co
Cm
Co Ci
(b2, 1) R
+ −
−
− ρ
Fig. 6. The level curves det(A3) = 0. The dashed lines correspond to the boundaries of the regions Cm, Co and Ci.
R R
ρ ρ
p q
D2 D3
D1
D1
D4
D5
[a]
[b]
[c]
[e]
[g]
[h]
[i]
[j]
[a]
[b]
[d]
[f ]
R = ρ/3 R = 3ρ
R
ρ
D5
D6
@@
R = ρ/3 R = 3ρ
[a]
[b]
[c]
[d]
[e]
[f ]
[g]
[h]
[i]
[j]
Fig. 7. Geometric properties of the nested tetraheda in D. We use the notation [{a, b, c}, a ∩ b, a∩c] where {a, b, c} denotes that a, b, and c are the scale factors of the inner, medium and outer tetrahedra, respectively, a ∩ b denotes that the faces of the tetrahedron with scale factor a intersect with the ones of the scale factor b, and a∩c denotes that the faces of the tetrahedron with scale factor a does not intersect with the ones of the scale factor c. Using this notation we obtain ten different re- gions [a]: [{ρ, 1, R}, ρ∩1, ρ∩R], [b]: [{ρ, 1, R}, ρ ∩ 1, ρ∩R], [c]: [{ρ, 1, R}, ρ ∩ 1, ρ ∩ R], [d]: [{1, ρ, R}, ρ ∩ 1, ρ∩R], [e]: [{1, ρ, R}, ρ ∩ 1, ρ ∩ R], [f]: [{1, ρ, R}, ρ∩1, ρ∩R], [g]:
[{1, ρ, R}, ρ∩1, ρ ∩ R], [h]: [{1, R, ρ}, ρ ∩ 1, ρ ∩ R], [i]: [{1, R, ρ}, ρ∩1, ρ ∩ R], [j]:
[{1, R, ρ}, ρ∩1, ρ∩R]. These ten regions are limited by the dashed and continuous lines in the figures.
the inner, the medium and the outer one. Statement (b) can be proved by analyzing these geometric properties of the three nested tetrahedra on the regions Di for i = 1, . . . , 6 (see Figure 7 for more details).
Finally we analyze the solution of (9) at the points p and q. The numeri- cal values of p and q can be found by solving system det(A) = det(A2) = det(A3) = 0.
Let
A(ρ) =
1 f (1, ρ) ρ −1/(2ρ2)
.
We note that det(A(ρ)) = −f3(ρ), so det(A(ρ)) 6= 0 for all ρ 6= b1 and ρ 6= b2. In particular, det(A(p1)) 6= 0 and det(A(q1)) 6= 0.
Since det(A) = det(A3) = 0 and det(A(ρ)) 6= 0, system (9) is equivalent to
system
1 f (1, ρ) ρ −1/(2ρ2)
λ m
=
1/2 − Mg1(R)
−f(ρ, 1) − Mf(ρ, R)
.
and it has infinitely many solutions which are given by
λ = λ(M) = 1 det(A)
1/2 − Mg1(R) f (1, ρ)
−f(ρ, 1) − Mf(ρ, R) −1/(2ρ2)
= −4ρ2f (1, ρ)f (ρ, 1) − 1
2 (2ρ3f (1, ρ) + 1) − M2ρ2f (1, ρ)f (ρ, R) + g1(R) 2ρ3f (1, ρ) + 1 , m = m(M) = 1
det(A)
1 1/2 − Mg1(R) ρ −f(ρ, 1) − Mf(ρ, R)
(12)
=ρ2(ρ + 2f (ρ, 1))
2ρ3f (1, ρ) + 1 − M2ρ2(ρg1(R) − f(ρ, R)) 2ρ3f (1, ρ) + 1 .
This solution provides a central configuration of the 12–body problem when λ(M) > 0, m(M) > 0 and M > 0. Moreover m(M) = 0 when M = (ρ2(ρ + 2f (ρ, 1)))/(2ρ2(ρg1(R) − f(ρ, R))).
Evaluating (12) at the point p we get λ = 0.3706146625 · · ·+0.6060692429 . . . M, and m = −0.4001317738 · · ·+3.1936671053 . . . M. Moreover this solution pro- vides a central configuration of the 12–body problem when M > 0.1252891302 . . . . Finally evaluating (12) at the point q we get the solution λ = −0.0001803428746 · · ·+
0.00001606707675 . . . M, and m = −3.5811747685 · · ·+0.0001420840585 . . . M, which provides a central configuration of the 12–body problem when M >
25204.620455 . . . . This completes the proof of statement (c).
3.1 The functions m(ρ, R) and M(ρ, R)
In this section we analyze some of the properties of the functions m(ρ, R) and M(ρ, R) on the regions Di for i = 1, . . . , 6.
After some tedious computations we can prove that
R→∞lim M(ρ, R) =
+∞ if ρ ∈ K1,
−∞ if ρ ∈ K2,
